Approximaths

Approximaths

  • Classic Summation Axioms

    So far, we’ve seen several ways of summing series (the usual method, Euler, Borel, generic, Borel-Écalle and Zeta summation). All of these methods fulfill all three properties — except Zeta summation, which fulfills none of them.

    \displaystyle \mathcal{S}(a_0+a_1+a_2+ ...) = \lim_{n \to \infty} \sum_{k =0}^{n}a_k \quad\text{(Regularity)}
    \displaystyle \mathcal{S}(\sum(\alpha a_n) + \sum(\beta b_n)) = \alpha \mathcal{S}(\sum a_n) + \beta \mathcal{S}(\sum b_n) \quad\text{(Linearity)}
    \displaystyle \mathcal{S}(a_0+a_1+a_2+ ...) = a_0 + \mathcal{S}(a_1+a_2+ ...) \quad\text{(Stability)}

    Concerning the Zeta summation we should restrain our enthusiasm by saying that a certain meromorphic complex function called \zeta which have the value \zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s} for all Re(s)>1 can be defined on the whole complex plane except at 1, in such a way that, \zeta(0) = -\frac{1}{2}. This ‘summation’ does not rely on any of the properties cited above. So it’s important to be clear about the method we are using and what properties it fulfills.

  • Zeta summation VI

    If we want to calculate \zeta(-2) for example we need to have a proper value for \Gamma(s/2) in for Re(s) \leq 0. The problem is that the Bernoulli representation of the Gamma function presented above:

    \displaystyle \Gamma(s) = \int_{0}^{\infty} e^{-t} t^{s-1}\,dt

    is only valid for Re(s) > 0. We have to provide another representation of the Gamma function for negative values of s:

    \Gamma(s) = \int_{0}^{\infty} e^{-t}\, t^{s-1}\, dt
    = \int_{1}^{\infty} e^{-t}\, t^{s-1}\, dt + \int_{0}^{1} e^{-t}\, t^{s-1}\, dt
    = \int_{1}^{\infty} e^{-t}\, t^{s-1}\, dt + \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} t^{n+s-1}\, dt
    = \int_{1}^{\infty} e^{-t}\, t^{s-1}\, dt + \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_{0}^{1} t^{n+s-1}\, dt
    = \int_{1}^{\infty} e^{-t}\, t^{s-1}\, dt + \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \left[ \frac{t^{n+s}}{n+s} \right]_{0}^{1}
    = \int_{1}^{\infty} e^{-t}\, t^{s-1}\, dt + \sum_{n=0}^{\infty} \frac{(-1)^n}{n!\,(n+s)}

    The sum \sum_{n=0}^{\infty} \frac{(-1)^n}{n!(n+s)} exists for negative s except for s= -1,-2,-3,... This implies that \frac{1}{\Gamma(s/2)} is entire with simple zeros at s= -2,-4,-6,.... The simple pole of \xi(s) at zero is cancelled by the corresponding zero of \frac{1}{\Gamma(s/2)}. As a consequence the only singularity of \zeta(s) is a single pole at s=1. Since

    \displaystyle \zeta(s) = \pi^{s/2}\frac{\xi(s)}{\Gamma(s/2)}

    and \frac{1}{\Gamma(s/2)} is entire with simple zeros at s= -2,-4,-6,..., this implies that for example:

    \displaystyle \zeta(-2) = 0

    We have in general

    \displaystyle \zeta(-2n)=0

    The so-called “trivial zeros” of the Zeta function. They occur at the negative even integers.

  • Zeta summation V

    In a previous post we have computed the meromorphic continuation into the entire complex plane of the Zeta function:

    \displaystyle \zeta(s) = \pi^{s/2}\frac{\xi(s)}{\Gamma(s/2)}

    This analytic continuation allows us to sum the Zeta function in the whole complex plane. For special (‘known’) values of the Xi function we can calculate the Zeta function. For example for

    s = 2 we have \xi(2) = \frac{\pi}{6} and \Gamma(2/2) = \Gamma(1) = 1 and therefore:

    \displaystyle \zeta(2) = \frac{\pi^2}{6}
    \displaystyle \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}

    For s = 4 we have \xi(4) = \frac{\pi^2}{90} and \Gamma(4/2) = \Gamma(2) = 1:

    \displaystyle \zeta(4) = \frac{\pi^4}{90}
    \displaystyle \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + ... = \frac{\pi^4}{90}

    For s = 6 we have \xi(6) = \frac{2\pi^3}{945} and \Gamma(6/2) = \Gamma(3) = 2:

    \displaystyle \zeta(6) = \frac{\pi^6}{945}
    \displaystyle \frac{1}{1^6} + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + ... = \frac{\pi^6}{945}

    So, Zeta of all the positive even integers has the form

    \displaystyle\zeta(2n) = (-1)^{n+1} \frac{B_{2n} (2\pi)^{2n}}{2(2n)!}

    Where B_{2n} are Bernoulli numbers.

  • 1 + 2 + 3 + 4 + …

    To calculate the value of the Riemann zeta function at s = -1, we use the functional equation (presented here):

    \displaystyle \zeta(s) = \pi^{s/2} \frac{\xi(s)}{\Gamma(s/2)}

    The Dirichlet series for the Riemann zeta function is defined for \Re(s) > 1 as:

    \displaystyle \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}

    For s = -1, the formal series (which is divergent in the classical sense) is:

    \displaystyle \zeta(-1) = \sum_{n=1}^{\infty} \frac{1}{n^{-1}}= \sum_{n=1}^{\infty} n = 1 + 2 + 3 + 4 + \dots

    Substituting s = -1 into the formula, we obtain:

    \displaystyle \zeta(-1) = \pi^{-1/2} \frac{\xi(-1)}{\Gamma(-1/2)}

    Using the recurrence property of the Gamma function, \Gamma(z) = \frac{\Gamma(z+1)}{z}, and the fact that \Gamma(1/2) = \sqrt{\pi}:

    \displaystyle \Gamma(-1/2) = \frac{\Gamma(-1/2+1)}{-1/2} = \frac{\Gamma(1/2)}{-1/2} = -2\sqrt{\pi}

    We exploit the functional equation \xi(s) = \xi(1-s), which implies:

    \displaystyle \xi(-1) = \xi(1 - (-1)) = \xi(2)

    From the definition of \xi(s), we have:

    \displaystyle \xi(2) = \pi^{-2/2} \Gamma(2/2) \zeta(2)

    Using \Gamma(1) = 1 and the solution of the Basel problem presented in the previous post \zeta(2) = \frac{\pi^2}{6}:

    \displaystyle \xi(-1) = \pi^{-1} \cdot 1 \cdot \frac{\pi^2}{6} = \frac{\pi}{6}

    Combining these results into our original expression:

    \displaystyle \zeta(-1) =\frac{1}{\sqrt{\pi}} \cdot \frac{\frac{\pi}{6}}{-2\sqrt{\pi}}
    \displaystyle = \frac{\pi}{-12 \cdot (\sqrt{\pi} \cdot \sqrt{\pi})}
    \displaystyle = \frac{\pi}{-12\pi}
    \displaystyle \boxed{\zeta(-1) = -\frac{1}{12}}

    Though surprising at first, turns out to be extremely useful in physics — most notably in string theory and the calculation of the Casimir effect, where the infinite sum 1 + 2 + 3 + 4 + … = −1/12 naturally appears when regularizing the vacuum energy between two conducting plates.

  • Zeta summation: Basel problem

    For s= 2, the zeta series is:

    \displaystyle \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^{2}}= 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots

    We would like to evaluate this series using the representation presented the previous post in we obtain

    \displaystyle \zeta(s) = \pi^{s/2} \frac{\xi(s)}{\Gamma(s/2)}

    For s = 2,

    \displaystyle \zeta(2) = \pi^{2/2} \frac{\xi(2)}{\Gamma(2/2)} = \pi \, \xi(2)

    since \Gamma(1) = 1. We need to evaluate

    \displaystyle \xi(2) = \int_0^\infty \frac{\vartheta(u) - 1}{2} \, du.

    We split the integral as follows:

    \displaystyle \int_0^1 \frac{\vartheta(u) - 1}{2} \, du + \int_1^\infty \frac{\vartheta(u) - 1}{2} \, du

    For the second part, we make the change of variable u = 1/v, so du = -dv/v^2. Using the functional equation of the theta function presented in a previous post.

    \displaystyle \vartheta(u) = u^{-1/2} \vartheta(1/u), \qquad u>0.

    This can be rewritten as

    \displaystyle \vartheta(1/v) = \sqrt{v} \, \vartheta(v)

    one can show (after standard calculations) that the two parts combine to yield a convergent integral. The final evaluation, which relies on the Poisson summation formula applied to the Gaussian, gives

    \displaystyle \xi(2) = \frac{\pi}{6}

    We obtain

    \displaystyle \zeta(2) = \pi \cdot \xi(2) = \pi \cdot \frac{\pi}{6} = \frac{\pi^2}{6}

    Finally

    \displaystyle 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots = \frac{\pi^2}{6}

    The Basel problem was sloved by Euler in 1735, who showed that the sum equals \displaystyle \frac{\pi^2}{6}.

  • Zeta summation IV

    In the previous post we derived the functional equation of the theta function

    \displaystyle \vartheta(t) = t^{-1/2} \vartheta(1/t), \qquad t>0.

    With t > 0. We observe that:

    \displaystyle \vartheta(t) \leq C t^{-1/2} \text{ as } t \to 0

    note that

    \displaystyle \sum_{n \geq 1}e^{-\pi n^2 t} \leq \sum_{n \geq 1}e^{-\pi n t} \leq Ce^{-\pi t}, t \geq 1

    Remember the definition of the Gamma function. For s >0 the Gamma function is defined by:

    \displaystyle \Gamma(s) = \int_{0}^{\infty} e^{-t} t^{s-1}\,dt
    \displaystyle \Gamma(s/2) = \int_{0}^{\infty} e^{-t} t^{\frac{s}{2}-1}\,dt

    Define u:= \frac{t}{\pi n^2} \implies du = \frac{1}{\pi n^2}\,dt and dt = \pi n^2 \,du

    \displaystyle \Gamma(s/2) = \int_{0}^{\infty} e^{-\pi n^2 u} (\pi n^2 u)^{\frac{s}{2}-1} \pi n^2\,du
    \displaystyle \Gamma(s/2) = \pi^{s/2} n^{s}\int_{0}^{\infty} e^{-\pi n^2 u} u^{\frac{s}{2}-1}\,du
    \displaystyle \pi^{-s/2} \Gamma(s/2) n^{-s} = \int_{0}^{\infty} e^{-\pi n^2 u} u^{\frac{s}{2}-1}\,du \qquad (n \geq 1)

    Observe that for the \vartheta(t) defined above:

    \displaystyle \vartheta(u) = \sum_{n \in \mathbb{Z}} e^{-\pi n^2 u}
    \displaystyle = 2 \sum_{n = 1}^{\infty} e^{-\pi n^2 u} + 1
    \displaystyle \frac{\vartheta(u) -1}{2} = \sum_{n = 1}^{\infty} e^{-\pi n^2 u}

    It follows that:

    \displaystyle \int_{0}^{\infty} u^{\frac{s}{2}-1} \biggl(\frac{\vartheta(u) -1}{2}\biggr) \,du = \int_{0}^{\infty} u^{\frac{s}{2}-1} \sum_{n = 1}^{ \infty} e^{-\pi n^2 u} \,du
    \displaystyle = \sum_{n = 1}^{\infty} \int_{0}^{\infty} u^{\frac{s}{2}-1} e^{-\pi n^2 u}\,du
    \displaystyle = \sum_{n = 1}^{\infty} \int_{0}^{\infty} e^{-\pi n^2 u} u^{\frac{s}{2}-1} \,du
    \displaystyle = \sum_{n = 1}^{\infty} \pi^{-s/2} \Gamma(s/2) n^{-s}
    \displaystyle = \pi^{-s/2} \Gamma(s/2) \sum_{n = 1}^{\infty} n^{-s}
    \displaystyle = \pi^{-s/2} \Gamma(s/2) \zeta(s)

    Therefore:

    \displaystyle \zeta(s) = \frac{\int_{0}^{\infty} u^{\frac{s}{2}-1} \biggl(\frac{\vartheta(u) -1}{2}\biggr) \,du}{\pi^{-s/2} \Gamma(s/2) }

    The define the Xi function as :

    \displaystyle \xi(s) := \int_{0}^{\infty} u^{\frac{s}{2}-1} \biggl(\frac{\vartheta(u) -1}{2}\biggr) \,du

    We finally have

    \boxed{\displaystyle \zeta(s) = \pi^{s/2}\frac{\xi(s)}{\Gamma(s/2)}}

    which is the meromorphic continuation into the complex plane of the Zeta function with a simple pole at s = 1.

  • Zeta summation III

    Recall that in the previous section our goal was to analytically extend the Riemann zeta function.

    To this end, we showed that the Gaussian function e^{-\pi x^2} is invariant under the Fourier transform:

    \displaystyle \mathcal{F}(e^{-\pi x^2})(\xi) = \int_{-\infty}^{\infty} e^{-\pi x^2} e^{-2\pi i x \xi}\,dx = e^{-\pi \xi^2}.

    We recall that the Fourier transform is defined by

    \displaystyle \mathcal{F}(f)(\xi) = \int_{-\infty}^{\infty} f(x)\,e^{-2\pi i x \xi}\,dx.

    For a>0, a change of variables shows that

    \displaystyle \mathcal{F}(f(ax))(\xi) = \frac{1}{a}\,\mathcal{F}\!\left(f\right)\!\left(\frac{\xi}{a}\right).

    Applying this to the Gaussian f(x)=e^{-\pi x^2} with a=\sqrt{t}, we obtain

    \displaystyle \mathcal{F}\!\left(e^{-\pi t x^2}\right)(\xi) = t^{-1/2} e^{-\pi \xi^2/t}.

    Both functions belong to the class \mathcal{S} of functions (The Schwartz class consists of infinitely differentiable functions on \mathbb{R}^n that, together with all their derivatives, decay faster than any polynomial at infinity (rapidly decreasing functions)) and Poisson summation applies:

    \displaystyle \sum_{n\in\mathbb{Z}} f(n) = \sum_{n\in\mathbb{Z}} \mathcal{F}(f)(n).

    In the present case, this yields

    \displaystyle \sum_{n\in\mathbb{Z}} e^{-\pi t n^2} = \sum_{n\in\mathbb{Z}} t^{-1/2} e^{-\pi n^2/t}.

    The left-hand side defines the theta function

    \displaystyle \vartheta(t) := \sum_{n\in\mathbb{Z}} e^{-\pi t n^2}.

    We immediately obtain the functional equation

    \displaystyle \boxed{\vartheta(t) = t^{-1/2} \vartheta(1/t),\qquad t>0.}

  • Zeta summation II

    Now look at:

    \displaystyle f(z) = e^{-\pi z^2}

    where z \in \mathbb{C}. This function is an entire function (differentiable in the entire complex plane). Since this function is entire the Cauchy’s integral theorem applies:

    \displaystyle \int_{\gamma_1} f(z)\,dz = 0

    Where \gamma_1 is any closed contour in the complex plane. Let us consider the contour presented in the figure below:

    \displaystyle \int_{-R}^{R} f(z) \,dz + \int_{R}^{R+ i \xi} f(z) \,dz + \int_{R+ i \xi}^{-R+ i \xi} f(z) \,dz + \int_{-R+ i \xi}^{-R} f(z) \,dz = 0

    For the first term of the integral we set: z(t) = t.
    For the second term: z(t) = R + it.
    For the third term: z(t) = t + i \xi.
    For the fourth term: z(t) = -R + it.
    The integral becomes:

    \displaystyle \int_{-R}^{R} f(t) \,dt + \int_{R}^{R+ i \xi} f(R + it) i\,dt + \int_{R+ i \xi}^{-R+ i \xi} f(t + i \xi) \,dt + \int_{-R+ i \xi}^{-R} f(-R + it) i\,dt =
    \displaystyle \int_{-R}^{R} f(t) \,dt + \int_{0}^{\xi} f(R + it) i\,dt + \int_{R}^{-R} f(t + i \xi) \,dt + \int_{\xi}^{0} f(-R + it) i\,dt =
    \displaystyle \int_{-R}^{R} e^{-\pi t^2} \,dt + \int_{0}^{\xi} e^{-\pi (R + it)^2} i\,dt + \int_{R}^{-R} e^{-\pi (t + i \xi)^2} \,dt + \int_{\xi}^{0} e^{-\pi (-R + it)^2} i\,dt = 0

    Observe that the last term is equal to (using u = -t):

    \displaystyle \int_{\xi}^{0} e^{-\pi (-R + it)^2} i\,dt = -\int_{-\xi}^{0} e^{-\pi (-R - iu)^2} i\,du

    The term (-R - iu)^2 can be written (R + iu)^2:

    \displaystyle \int_{\xi}^{0} e^{-\pi (-R + it)^2} i\,dt = -\int_{-\xi}^{0} e^{-\pi (R + iu)^2} i\,du

    now:

    \displaystyle -\int_{-\xi}^{0} e^{-\pi (R + iu)^2} i\,du = - \left( -\int_{0}^{-\xi} e^{-\pi (R + iu)^2} i\,du \right) = \int_{0}^{-\xi} e^{-\pi (R + iu)^2} i\,du

    The second and fourth term cancel and the equation above becomes:

    \displaystyle \int_{-R}^{R} e^{-\pi t^2} dt + \int_{R}^{-R} e^{-\pi (t + i \xi)^2}\,dt = 0 \\ \int_{-R}^{R} e^{-\pi t^2} dt = - \int_{R}^{-R} e^{-\pi (t + i \xi)^2}\,dt \\ \int_{-R}^{R} e^{-\pi t^2} dt = \int_{-R}^{R} e^{-\pi (t^2 + 2 t i \xi - \xi^2)}\,dt \\ \int_{-R}^{R} e^{-\pi t^2} dt = e^{\pi \xi^2} \int_{-R}^{R} e^{-\pi t^2} e^{-2\pi i t \xi}\,dt \\

    Using

    \displaystyle \lim_{R \to \infty} \int_{-R}^{R} e^{-\pi t^2}\,dt = 1

    Therefore, if \xi \in \mathbb{R}:

    \displaystyle e^{-\pi \xi^2} = \int_{-\infty}^{\infty} e^{-\pi x^2} e^{-2\pi i x \xi} \,dx

    We conclude from this post and the previous post that the Gaussian function e^{-\pi \xi^2} is its own Fourier transform.

  • Zeta summation I

    The Zeta function is defined in the complex-plan for Re(z) > 1 as:

    \displaystyle \zeta(z)=\frac{1}{1^z}+\frac{1}{2^z}+\frac{1}{3^z}+\frac{1}{4^z}+...=\sum_{n=1}^{\infty}\frac{1}{n^z}

    For example:

    \displaystyle \zeta(2)=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...=\frac{\pi^2}{6}

    If we would like to use the Zeta function in order to perform a summation with Re(z)\leq 1 we should use the analytic continuation of the Zeta function.

    If f(z) is single-valued in some region of the complex-plan, the derivative of f(z) is defined as:

    \displaystyle f'(z)=\lim_{\Delta z\to0}\frac{f(z+\Delta z)-f(z)}{\Delta z}

    If this limit exists we say that f(z) is differentiable at z. If the derivative exists in all points of a region of the complex-plan we say that f(z) is analytic in this region. If the function is analytic inside some circle of convergence C_0 it can be represented by the Taylor series:

    \displaystyle a_0+a_1(z-z_0)+a_2(z-z_0)^2+...

    where z_0 is the center of the circle C_0. By choosing a new point z_1 in C_0 which is the center of a new circle C_1 the function can be represented by the Taylor series:

    \displaystyle b_0+b_1(z-z_1)+b_2(z-z_1)^2+...

    The figure above illustrates analytic continuation. I like to imagine analytic functions as exquisite porcelain—delicately fragile, yet rigid, able to be extended through analytic continuation.

    In order to analytically continue the Zeta function we first consider the following function:

    \displaystyle f(\xi)=e^{-\pi \xi^2}

    where \xi\in\mathbb{R}. We would like to show that this function is equivalent to its Fourier transform

    \displaystyle f(\xi)=e^{-\pi \xi^2}=\int_{-\infty}^{\infty}e^{-\pi x^2}e^{-2\pi i x\xi}\,dx

    if \xi=0:

    \displaystyle f(0)=1=\int_{-\infty}^{\infty}e^{-\pi x^2}\,dx

    if \xi<0 :

    \displaystyle \int_{-\infty}^{\infty} e^{-\pi x^2} e^{-2\pi i x \xi}\,dx = \int_{-\infty}^{\infty} e^{-\pi u^2} e^{2\pi i u (-\xi)}\,du = \int_{-\infty}^{\infty} e^{-\pi u^2} e^{-2\pi i u |\xi|}\,du = e^{-\pi \xi^2}

    The demonstration will be completed in the next post.

    I wish you a merry Christmas 2025 !

  • Anharmonic oscillator II

    Remember that the perturbative series for the anharmonic oscillator is

    Anharmonic oscillator I

    \displaystyle E_0(\epsilon) = \frac{1}{2} + \frac{3}{4}\epsilon - \frac{21}{8}\epsilon^2 + \frac{333}{16}\epsilon^3 + \mathcal{O}(\epsilon^4)

    We used Padé approximants to compute the ground state energy E_0(\epsilon). Now we aim to calculate E_0(\epsilon) using Borel summation. The (formal) Borel sum is given by

    \displaystyle \int_{0}^{\infty} e^{-t} \sum_{n=0}^\infty \frac{a_n}{n!} (\epsilon t)^n \, dt.

    Using the first three coefficients of the perturbative series, the truncated Borel transform is approximated by

    \displaystyle \mathcal{B}E_0(t) \approx \frac{1}{2} + \frac{3}{4} t - \frac{21}{16} t^2,

    so the truncated Borel sum writes

    \displaystyle E_0(\epsilon) \approx \int_0^\infty e^{-t} \left(\frac{1}{2} + \frac{3}{4} \epsilon t - \frac{21}{16} \epsilon^2 t^2 \right) dt = \frac{1}{2} + \frac{3}{4} \epsilon - \frac{21}{8} \epsilon^2.

    since

    \displaystyle \int_0^\infty \frac{1}{2} e^{-t} \, dt = \frac{1}{2}
    \displaystyle \int_0^\infty e^{-t} \cdot \frac{3}{4} \epsilon \, t \, dt = \frac{3}{4} \epsilon
    \displaystyle \int_0^\infty e^{-t} \cdot \left( -\frac{21}{16} \epsilon^2 t^2 \right) dt = -\frac{21}{8} \epsilon^2

    This doesn’t really seem like progress, since the first terms of the Borel sum are identical to those of the perturbative expansion. The perturbative expansion E_0(\epsilon) = \sum_{n=0}^\infty a_n \epsilon^n above has coefficients that grow factorially. Bender and Wu showed that for large n,

    \displaystyle a_n \sim -r\, (-1)^n \left(\frac{3}{2}\right)^n \Gamma\!\left(n+\tfrac12\right), \qquad r>0.

    Since

    \displaystyle \Gamma\!\left(n+\tfrac12\right) \sim n!\, n^{-1/2},
    formule avec n! en rouge et gras

    thus the series diverges for all \epsilon \neq 0.

    Borel summation improves convergence by dividing out this factorial growth.

    In summary, even with just a few terms, Borel summation correctly recovers the perturbative results for the anharmonic oscillator then It turns the divergent series into a well-defined and useful result.